Page:Optics.djvu/86

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87. ⁠Now in the first place it will be immediately seen that this expression gives the principal focal distance, which we will call $F$, by leaving out the last term, which is equivalent to making ${\frac {1}{\Delta }}=0$, or $\Delta$ infinite: we have thus

${\frac {1}{F}}=(m-1)\left\{{\frac {1}{r}}-{\frac {1}{r'}}\right\}$^[1]

and then,

${\frac {1}{\Delta ^{n}}}={\frac {1}{F}}+{\frac {1}{\Delta }}.$

It appears from the former of these that $F$ is positive or negative according as ${\frac {1}{r}}-{\frac {1}{r'}}$ is so: let us examine what sign this is affected with in different cases.

In the concavo-convex lens placed as in Fig. 86, $r<r'$ and $F$ is positive.

When this lens is turned the contrary way, $r>r'$, but they are both negative, we have then

${\frac {1}{F}}=(m-1)\left\{{\frac {1}{r'}}-{\frac {1}{r}}\right\},$

and $F$ is positive as before.

In the meniscus, either $r>r',$ both being positive, and then

${\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r'}}-{\frac {1}{r}}\right\},$

or $r<r'$, and both are negative: so that

${\frac {1}{F}}=-(m-1)\left\{{\frac {1}{r}}-{\frac {1}{r'}}\right\}.$

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1. It is often found convenient to put some symbol such as ${\frac {1}{\rho }}$ for ${\frac {1}{r}}-{\frac {1}{r'}},$ which gives ${\frac {1}{F}}={\frac {m-1}{\rho }}$, or $F={\frac {\rho }{m-1}}$. When the radii are equal in a double concave or convex lens $\rho ={\frac {r}{2}}$.